We are given a triangle ABC with sides AC = 12, BC = 10, and we are asked to find angle B. Side AB is labeled 's', but its value is not given and is not needed to find angle B.
Since we have two sides and the angle opposite one of them is the angle we want to find, we can use the Law of Cosines. The Law of Cosines states that for any triangle with sides a, b, and c and angles A, B, and C (opposite sides a, b, and c respectively), \(c^2 = a^2 + b^2 - 2ab\cos(C)\). In our case, we have \(AC^2 = BC^2 + AB^2 - 2(BC)(AB)\cos(B)\). We can rearrange the formula to solve for cos(B): \(\cos(B) = \frac{BC^2 + AB^2 - AC^2}{2(BC)(AB)}\). Plugging in the given values, \(AC = 12\) and \(BC = 10\), we need AB to use the equation.
However, another form of the law of cosines is \(AC^2 = BC^2 + AB^2 - 2(BC)(AB)cos(C)\). In this case, let a = BC = 10, b = AB = s and c = AC = 12. Also, let C = angle C. Since we're trying to solve for angle B, we have \(b^2 = a^2 + c^2 - 2ac \cos(B)\). In our case, \(AB^2 = BC^2 + AC^2 - 2(BC)(AC) \cos(\angle B)\), or \(12^2 = 10^2 + s^2 - 2_10_s \cos(C)\)
Using the Law of Cosines with \(a=10, c=12\) and angle B, we have
\(12^2 = 10^2 + s^2 - 2(10)(s)\cos(B)\)
Instead, use \(b = s, a = 10, c = 12\), to find angle B.
\(144 = 100 + s^2 - 20s \cos(B)\)
We don't need side \(s\) since we are looking for the angle opposite to side 12. Let a = 10, b = s, c= 12, so C is the angle opposite side c = 12. Thus, C is the angle at B.
\(c^2 = a^2 + b^2 - 2ab\cos(C)\)
\(12^2 = 10^2 + s^2 - 2(10)(s)\cos(B)\)
\(144 = 100 + s^2 - 20s \cos(B)\)
However, if we use a = BC = 10, b = AC = 12, and c = AB, with the angle opposite AB being C = Angle C at C, we have \(s^2 = 10^2 + 12^2 - 2_10_12\cos(C)\)
This formula cannot help find angle B.
The correct form to use is \(AC^2 = BC^2 + AB^2 - 2(BC)(AB) \cos( \angle B)\). Here, we'll let \(AC = b = 12, BC = a = 10, AB = c = s\) and angle B is opposite side AC. So,
\(12^2 = 10^2 + s^2 - 2(10)(s) \cos(B)\). The \(s^2\) and the \(-20s\cos(B)\) cannot be isolated, so let's try \(AB^2 = BC^2 + AC^2 - 2(BC)(AC) \cos(\angle C)\).
We have \(s^2 = 10^2 + 12^2 - 2(10)(12) \cos(\angle C)\).
\(b^2 = a^2 + c^2 - 2ac\cos(B)\)
\(12^2 = 10^2 + s^2 - 2(10)(s) \cos(B)\)
Using the law of cosines: \(b^2 = a^2 + c^2 - 2ac \cos{B}\), we let a = 10, b = 12, c = s, and the angle we seek is B.
\(12^2 = 10^2 + s^2 - 2(10)(s) \cos(B)\). We have 2 unknowns.
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos(\angle B)\). However, we don't know AB = s.
\(b^2 = a^2 + c^2 - 2ac \cos(B)\)
\(144 = 100 + s^2 - 20s\cos(B)\)
\\(\boxed{38.6^{\circ}}\\)
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